Given a tree, calculate the average distance between two vertices in the tree. For example, the average distance between two vertices in the following tree is (d 01 + d 02 + d 03 + d 04 + d 12 +d 13 +d 14 +d 23 +d 24 +d 34)/10 = (6+3+7+9+9+13+15+10+12+2)/10 = 8.6. 

Input

On the first line an integer t (1 <= t <= 100): the number of test cases. Then for each test case: 

One line with an integer n (2 <= n <= 10 000): the number of nodes in the tree. The nodes are numbered from 0 to n - 1. 

n - 1 lines, each with three integers a (0 <= a < n), b (0 <= b < n) and d (1 <= d <= 1 000). There is an edge between the nodes with numbers a and b of length d. The resulting graph will be a tree. 
 

Output

For each testcase: 

One line with the average distance between two vertices. This value should have either an absolute or a relative error of at most 10 -6 
 

Sample Input

1
5
0 1 6
0 2 3
0 3 7
3 4 2

Sample Output

8.6

解題報告：

就是求出樹上任意兩點間的距離的和，再求平均值。這樣只要知道每個邊走過幾次，用次數乘上這條邊的權值，再把所有邊加和，除以路徑數就可以。首先，關於路徑數，我覺得就像一個有 N個點的完備圖，這個圖中邊的數量，就是這N個點形成的樹，樹中任意兩點組成的邊的數量

ac程式碼：

#include<stdio.h>
#include<iostream>
#include<map>
#include<vector>
#include<string.h>
#define ll long long
#define mod 1000000007
using namespace std;
const int MAXN=2e5+10;
struct edge{
	int y;
	int w;
	edge(){}
	edge(int y,int w):y(y),w(w){}
}; 
int n;
vector<edge> vv[MAXN];
ll ans;
ll dp[MAXN];
void dfs(int x,int f)
{
	int ed=vv[x].size();
	for(int i=0;i<ed;i++) 
	{
		edge v=vv[x][i];
		if(v.y==f)
		continue;
		dfs(v.y,x);
		dp[x]+=dp[v.y];
		ans+=dp[v.y]*(n-dp[v.y])*v.w;
	}
}
int main()
{
	int t;
	scanf("%d",&t);
	while(t--)
	{
		scanf("%d",&n);
		int u,v,w; 
		ans=0;
		for(int i=0;i<n;i++)
		{	
			vv[i].clear();
			dp[i]=1;
		}
		for(int i=0;i<n-1;i++)
		{
			scanf("%d%d%d",&u,&v,&w);
			vv[u].push_back(edge(v,w));
			vv[v].push_back(edge(u,w));
		}
		dfs(0,-1); 
		printf("%lf\n",ans*2.0/n/(n-1));
	}
	return 0; 
};